Splitting an infinite set into two equal infinite subsets.

So if A is an infinite set, we know that |A|+|A|=|A|. But are we allowed to go backwards, i.e. divide A into two disjoint subsets B and C such that A = B U C, and |B|=|C|=|A|. For the integers and for the reals, this is clear, e.g. R = (-infinity,0) U [0, infinity), each with cardinality c. But are we allowed to do this for any infinite set of larger cardinality? I'm sure the answer is yes, but how do we go about proving this? Do we have to use the Axiom of Choice and transfinite induction? How complicated will the proof be?

How about this: Well-order A. Put the first element in the left bin, the second in the right bin,...alternatingly put the elements of A
in the left bin and right bin. By transfinite induction, this can be done for all elements of A. Then each bin has |A| elements.

Let A={a_i| i in I}. Well-order I. Suppose all elements a_i can be placed alternatingly in bin B and bin C for i < j. Then a_j can be placed in B if the previous was placed in C (and vice versa). So by transfinite induction, this can be done for all i in I, so A=BUC. We have B and C disjoint, and |B|=|C|. Then
|A|=|BUC|=|B|+|C|=2|B|=|B|=|C|. Sounds good?

Wait, wait, do you want A = B U C, or |A| = |B U C|? Because if it's the latter, all you have to do is notice that |A| = | {0}xA U {1}xA |

Partition A into B U C, such that |B|=|C|=|A|. Actually, we should be able to generalize this to k partitions of A such that each partition has cardinatility |A|, and k is any (infinite) number less than |A|, right?

I think you're misunderstanding the question: he's asking if, for an infinite set A, the the system of equations

A = B U C
|A| = |B|
|A| = |C|

has a solution for B and C.

i knew what he meant and what i said stands. c + c = c, and [itex]\aleph_0 + \aleph_0 =\aleph_0[/itex] if the sets are disjoint and if they're not they can be made so by using dragonfall's method. hence c = c + c and [itex]\aleph_0 = \aleph_0 + \aleph_0 [/itex].

So if A is an infinite set, we know that |A|+|A|=|A|. But are we allowed to go backwards, i.e. divide A into two disjoint subsets B and C such that A = B U C, and |B|=|C|=|A|. For the integers and for the reals, this is clear, e.g. R = (-infinity,0) U [0, infinity), each with cardinality c. But are we allowed to do this for any infinite set of larger cardinality? I'm sure the answer is yes, but how do we go about proving this? Do we have to use the Axiom of Choice and transfinite induction? How complicated will the proof be?

if the question is this :

|A| + |A| = |A| :1
|B|=|C|=|A| :2

|B|+|C| [tex]\stackrel{?}{=}[/tex] |A|

how is it not immediate that by 2
(|A|=|B|) + (|A|=|C|) = |B|+|C| = |A| ?

I'm not trying to prove that |B|+|C| = |A|. That follows immediately from |A| + |A| = |A|. I'm trying to prove that B and C exist such that B and C partition A and each have cardinality |A|. But I think I've already proven it now, and am wondering about the infinite partitioning case.

again huh? what do you mean by partition? here are three theorems from my set theory book that might be what you're looking for :

countable union of countable sets is countable
any infinite set can written as the union of disjoint countably infinite sets.
any infinite set A can be written as a disjoint union A = B U C where A,B, and C have the same cardinality.

That answers the question.
Starting with the knowledge that |A|=|A|+|A|, no choice is required, it is just a matter of applying the definitions.

More precisely, |A|=|A|+|A| just means that there are two disjoint sets B,C such that |B|=|C|=|A| and |A|=|B U C|. By definition B U C can be put into 1-1 correspondence with A. Let B' be the image of B and C' be the image of C. Then A = B' U C' and |B'|=|C'|=|A|.